# Thread: Second order DE help

1. ## Second order DE help

Hi,

I have the equation:
$\displaystyle \begin{cases}y^{\prime\prime} + 3y^{\prime} + 2y = sin~x \\ y(0) = 1 \\ y^{\prime}(0) = 1 \end{cases}$

The auxiliary equation $\displaystyle t^2 + 3t + 2 = 0$ has the solutions
$\displaystyle x_{1}=-1~~ x_{2}=-2$

So the general solution would look like this
$\displaystyle Ae^{-t}+Be^{-2t}$

Now, let $\displaystyle \begin{cases}y = A~Cos(t)+B~Sin(t) \\ y^{\prime} = -A~Sin(t)+B~Cos(t) \\ y^{\prime\prime} = -A~Cos(t)-B~Sin(t) \end{cases}$

gives $\displaystyle -A~Cos(t)-B~Sin(t) + 3\cdot(-A~Sin(t)+B~Cos(t)) + 2\cdot(A~Cos(t)+B~Sin(t))$

Can be simplified down to $\displaystyle 3A~Cos(t)-3B~Sin(t) + ASin(t) +B~Cos(t)$

This is where i am stuck, how do you combine this with the general solution?

//Thanks

2. Well, you know that $\displaystyle 3A+B=0$ since there are no $\displaystyle cos x$ terms, and $\displaystyle -3B+A=0$ since there is exactly one $\displaystyle sin x$ term.

As a sidebar, it's easier to answer things like this clearly when you use standard notations. Your initial equation indicates $\displaystyle x$ as the independent variable, and $\displaystyle \lambda$ is what is typically used for calculating the characteristic equations.

3. ## That doesn´t look right

If $\displaystyle 3A + B = 0$ (1) $\displaystyle -> = B + 3A = 0$
and $\displaystyle B - 3A = 0$ (2) $\displaystyle -> = B - 3A = 0$ then A must be 0

From (1) -> $\displaystyle B = -3A$
Substituting into (2) $\displaystyle -> -3A - 3A = 0 -> -6A = 0$ then A = 0

And from (2) $\displaystyle -> B = 3A$ substituting into $\displaystyle (1) -> B + B = 0$ then B must be 0

Since you have $\displaystyle Ae^-t + Be^-2t$ you would get $\displaystyle y;y`= 0$

4. ## Accually you can

Was the textbook answer -0.5e^(-t) + -0.5e^(-2t) -0.5cosx -0.5sinx???

5. Originally Posted by Henryt999
Was the textbook answer -0.5e^(-t) + -0.5e^(-2t) -0.5cosx -0.5sinx???
I don't know, i don't have the correct answer...

Anyway,

we have $\displaystyle \begin{cases}3A + B = 0 \\ -3B + A = 1\end{cases}$

A = 0.1
B = -0.3

So $\displaystyle y_p = 0.1Sin(t)-0.3Cos(t)$
But how do i pair this up with the general solution?

6. Originally Posted by Jones
I don't know, i don't have the correct answer...

Anyway,

we have $\displaystyle \begin{cases}3A + B = 0 \\ -3B + A = 1\end{cases}$

A = 0.1
B = -0.3

So $\displaystyle y_p = 0.1Sin(t)-0.3Cos(t)$
But how do i pair this up with the general solution?
Just add it to the general solution of the homogeneous equation!

If L(y) is a linear differential operator, then L(y+ y2)= L(y1)+ L(y2). If yh satifies L(yh)= 0 and yp satisfies L(yp)= f(x), then L(yh+ yp)= L(yh)+ L(yp)= 0+ f(x)= f(x). Further, if y is any solution to L(y)= f(x), then L(y- yp)= L(y)- L(yp)= f(x)- f(x)= 0 so that y- yp= yh, a solution to the homogeneous equation. The general solution to the entire equation is the the general solution to the homogeneous equation plus any solution to the entire equation. That's the whole point of this method!